$\begingroup$You may find expressions for the module for $a=0, a=\frac 12$ or $a=1$ here ($(6.1.29)$ and more) (btw your link shows rather a relation of the argument of zeta with $\zeta(1/2+it)$ : $\Gamma$ is the 'simpler' function!).$\endgroup$
– Raymond ManzoniDec 23 '12 at 9:16

$\begingroup$You are welcome ! (sorry for the typos 'module' $\to$ modulus and the MO link gives a relation of the argument of zeta with $\Gamma(1/2+it)$ : zeta is the complicated function and the relation just shows that its argument is simple since it may be written as a function of $\Gamma$ and other elementary functions). $\Gamma$ itself has only trivial closed forms as opposed to the derivative of $\log(\Gamma)$ i.e. the digamma or $\psi$ function that admits a closed form at every rational value.$\endgroup$
– Raymond ManzoniDec 23 '12 at 10:14

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$\begingroup$I don't know of an impossibility proof. Proofs exist that the gamma function can't be written as an elementary function but this doesn't exclude specific closed forms as at $\frac 12$.$\endgroup$
– Raymond ManzoniDec 23 '12 at 10:32

1 Answer
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The $\Gamma$ function lacks a closed form containing only elementary functions. Here are the equivalent formulas of the $\Gamma$ function however:

$$\Gamma(z)=\int_{0}^{+\infty}t^{z-1}e^{-t}dt$$
$$\Gamma(z)=\frac1z\prod_{n=1}^{\infty}\frac{(1+\frac1n)^z}{1+\frac zn}$$
$$\Gamma(z)=\frac{e^{-\gamma z}}{z}\prod_{n=1}^{\infty}\frac{e^{\frac zn}}{1+\frac zn}$$
where $\gamma$ is the Euler–Mascheroni constant. Of course we can relate the $\Gamma$ function with elementary functions via the following indentity:
$$\Gamma(1-z)\Gamma(z)=\frac{\pi}{\sin \pi z}$$
We also have Riemann's functional equation
$$\zeta(s)=2^s\pi^{s-1}\sin\frac{\pi s}2\Gamma(1-s)\zeta(1-s)$$
which relates the $\Gamma$ and the $\zeta$ functions.

If by closed form we mean an expression containing only elementary functions then no, $\Gamma$ has no such form. For more information read this